# Speed, momentum and total energy of a muon (SR)

• jasonchiang97
In summary, to calculate the speed, momentum, and total energy of a 200MeV muon with a mass of 106MeV/c2, we can use the equations E=γmc, K+mc2=E, γ=1/(1-β)1/2, and β=(v/c)2. After correctly expanding and rearranging, we get the expression √[c2(1-(K/mc2 + 1)2)]=v for the speed. By plugging in the values for the mass and kinetic energy of the muon, we can calculate the numerical value for v.
jasonchiang97

## Homework Statement

A muon has a mass of 106MeV/c2. Calculate the speed, momentum and total energy of a 200MeV muon(a muon with a kinetic energy of 200MeV).

E=γmc
K+mc2=E
γ=1/(1-β)1/2
β=(v/c)2

## The Attempt at a Solution

To solve for the speed I plugged E=γmc into K+mc2=E to obtain K/mc2 + 1 = γ. Expanding gamma and rearranging I get 1-(K/mc2+1)-1 = β. But plugging in the numbers on the left hand side gives me a zero as K<<c2 and I think that I went wrong somewhere.

jasonchiang97 said:

## Homework Equations

E=γmc
K+mc2=E
γ=1/(1-β)1/2
β=(v/c)2
The first, third, and fourth equations have errors.
[EDIT: Sorry. Your third and fourth equations are OK if you are defining the symbol β to equal (v/c)2. But it is more common to use β for v/c.]

3. The Attempt at a Solution
To solve for the speed I plugged E=γmc into K+mc2=E to obtain K/mc2 + 1 = γ.
E ≠ γmc. But your expression for γ looks correct.

Expanding gamma and rearranging I get 1-(K/mc2+1)-1 = β.
Check your work here, I don't think you got the correct expression for ##\beta##.

But plugging in the numbers on the left hand side gives me a zero as K<<c2
Did you mean to write K << mc2? If so, how did you come to this conclusion based on the information given in the problem?

Last edited:
Sorry line 1 was supposed to be E=γmc2 and sorry I skipped so many steps. I'll write down everything I did.

K=mc2(γ-1)
K/mc2 + 1 = 1/(√1-(v2/c2))
1-(v2/c2)=(K/mc2 + 1)2
1-(K/mc2 + 1)2=(v2/c2)
c2(1-(K/mc2 + 1)2)=v2
√[c2(1-(K/mc2 + 1)2)]=v

did I do anything wrong?

Did you mean to write K << mc2? If so, how did you come to this conclusion based on the information given in the problem?

Well given the numbers, c2 = 9 x 1016 and from the values of K and m given K/mc2 = 0

jasonchiang97 said:
Sorry line 1 was supposed to be E=γmc2 and sorry I skipped so many steps. I'll write down everything I did.

K=mc2(γ-1)
K/mc2 + 1 = 1/(√1-(v2/c2))
1-(v2/c2)=(K/mc2 + 1)2
...
did I do anything wrong?
Check going from 2nd to 3rd line

jasonchiang97 said:
Did you mean to write K << mc2? If so, how did you come to this conclusion based on the information given in the problem?

Well given the numbers, c2 = 9 x 1016 and from the values of K and m given K/mc2 = 0
Note the units in the value of the mass, m =106 MeV/c2.
So c2 cancels in mc2.

jasonchiang97 said:
Sorry line 1 was supposed to be E=γmc2 and sorry I skipped so many steps. I'll write down everything I did.

K=mc2(γ-1)
K/mc2 + 1 = 1/(√1-(v2/c2))
1-(v2/c2)=(K/mc2 + 1)-2
1-(K/mc2 + 1)-2=(v2/c2)
c2(1-(K/mc2 + 1)-2)=v2
√[c2(1-(K/mc2 + 1)-2)]=v

did I do anything wrong?

I keep making mistakes as I try to type it on the site sorry and thanks.

TSny said:
Note the units in the value of the mass, m =106 MeV/c2.
So c2 cancels in mc2.

Yea I know they cancel I mean when you put in the numbers you basically get 1/∞ for E/mc2

What numerical value do you get for mc2 (including units)?

(Your final expression for v in post #7 looks correct.)

Last edited:
rcordasco
TSny said:
What numerical value do you get for mc2 (including units)?

(Your final expression for v in post #7 looks correct.)

ohh I see the c2 cancels out. Thanks!

## 1. What is the speed of a muon in Special Relativity?

In Special Relativity, the speed of a muon is always less than the speed of light, which is approximately 3.0 x 10^8 meters per second. However, due to its high energy and short lifespan, a muon can travel at speeds close to the speed of light.

## 2. How is momentum calculated for a muon in Special Relativity?

In Special Relativity, the momentum of a muon can be calculated using the equation p = mv/√(1-v^2/c^2), where p is the momentum, m is the mass of the muon, v is its velocity, and c is the speed of light.

## 3. What is the relationship between speed and energy of a muon in Special Relativity?

In Special Relativity, the speed and energy of a muon are directly related. As the speed of the muon approaches the speed of light, its energy increases exponentially. This is described by the equation E = mc^2/√(1-v^2/c^2), where E is the energy, m is the mass of the muon, and c is the speed of light.

## 4. What is the total energy of a muon in Special Relativity?

In Special Relativity, the total energy of a muon includes both its rest energy (mc^2) and its kinetic energy (1/2mv^2), where m is the mass and v is the velocity of the muon. This can be represented by the equation E = mc^2 + 1/2mv^2.

## 5. How does the energy of a muon change with its relativistic speed?

In Special Relativity, the energy of a muon increases as its speed approaches the speed of light. This is due to the increase in kinetic energy as the muon's speed increases. However, as the muon's speed reaches the speed of light, it would require infinite energy to accelerate it further, making it impossible to reach the speed of light. This is described by the equation E = mc^2/√(1-v^2/c^2).

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